0
$\begingroup$

Let $R=\frac{\mathbb{Q}[t]}{(t^2-1)}$; trivially, $R$ is not an integral domain, since $(\overline{t-1})(\overline{t+1})=\overline{t^2-1}=\overline{0}$.

Is it possible to find $A,B,w \in R[x,y]$ such that the following three conditions are satisfied:

(i) $\operatorname{Jac}(A,B)\in \{1,-1\}$. (ii) $\operatorname{Jac}(A,w)=0$. (iii) $w \notin R[A]$.

Remarks:

  • Similar questions are: 1, 2, 3, 4; especially question 3.

  • Here we are not able to talk about normal or non-normal domains (in the simplest meaning), because now $R$ is not an integral domain.

  • Notice that $A=\overline{t−1}x+\overline{t}y$, $B=\overline{t}x+\overline{t+1}y$, $w=\overline{t+1}y$ do not satisfy all the three conditions; indeed, conditions (i) and (ii) are satisfied, but condition (iii) is not, since $R[A] \ni \overline{t+1}A=\overline{t+1}(\overline{t−1}x+\overline{t}y)= \overline{t^2-1}x+\overline{(t+1)t}y=\overline{0}x+\overline{t^2+t}y=\overline{1+t}y=w$.

Thank you very much!

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.